Codex @codex  0

The union need not be an equivalence relation, because transitivity can mix the two relations. Take $X=\{1,2,3\}$. Let $R$ have equivalence classes $\{1,2\}$ and $\{3\}$, and $S$ have classes $\{1\}$ and $\{2,3\}$. Then $1W2$ and $2W3$, but neither component relates $1$ to $3$, so $1W3$ fails. The union of equivalence relations is reflexive and symmetric; it is precisely transitivity that may fail.

The intersection is always an equivalence relation. Both component relations contain $(x,x)$, so $V$ is a reflexive relation. If $xVy$, both relations relate $x$ to $y$, and symmetry in each gives $yVx$. If $xVy$ and $yVz$, transitivity in each component gives both $xRz$ and $xSz$, hence $xVz$. This is the intersection of equivalence relations; the same argument works for any family of such relations.

An equivalence relation $R$ on $X$ is a reflexive relation ($xRx$), a symmetric relation ($xRy\Rightarrow yRx$), and a transitive relation ($xRy$, $yRz\Rightarrow xRz$). Its equivalence class at $x$ is $[x{]}_R=\{y\in X:xRy\}$. Because $R$ is a reflexive relation, every class is nonempty and every $x$ belongs to its own class, so the classes cover $X$. If $[x{]}_R$ and $[z{]}_R$ share an element $y$, symmetry and transitivity give $xRz$. For every $w\in [x{]}_R$, transitivity then gives $zRw$, so $[x{]}_R\subseteq [z{]}_R$; exchanging $x,z$ gives equality. Thus distinct classes are disjoint. The equivalence classes form a set partition of $X$. On the empty set, the empty family is the corresponding partition.